Question
Evaluate $\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t d t$
✓
Answer
Let $I=\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t d t$.
Consider $\int \sin ^{3} 2 t \cos 2 t d t$
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = $\frac{1}{2}$du
So $\int \sin ^{3} 2 t \cos 2 t d t=\frac{1}{2} \int u^{3} d u$
= $\frac{1}{8}\left[u^{4}\right]=\frac{1}{8} \sin ^{4} 2 t=\mathrm{F}(t)$
Therefore, by the second fundamental theorem of integral calculus
$I=F\left(\frac{\pi}{4}\right)-F(0)=\frac{1}{8}\left[\sin ^{4} \frac{\pi}{2}-\sin ^{4} 0\right]=\frac{1}{8}$
Consider $\int \sin ^{3} 2 t \cos 2 t d t$
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = $\frac{1}{2}$du
So $\int \sin ^{3} 2 t \cos 2 t d t=\frac{1}{2} \int u^{3} d u$
= $\frac{1}{8}\left[u^{4}\right]=\frac{1}{8} \sin ^{4} 2 t=\mathrm{F}(t)$
Therefore, by the second fundamental theorem of integral calculus
$I=F\left(\frac{\pi}{4}\right)-F(0)=\frac{1}{8}\left[\sin ^{4} \frac{\pi}{2}-\sin ^{4} 0\right]=\frac{1}{8}$
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